Vector Fields Multivariable Functions

Multivariable Functions Suppose f (x, y ) = y − x. Then, we could come up with a table of input values to compute different output values of f (x, y ). x y 3 2 1 0 -1 -2 -3 -3 -2 -1 0 1 2 3 6 5 4 3 2 1 0 5 4 3 2 1 0 -1 4 3 2 1 0 -1 -2 3 2 1 0 -1 -2 -3 2 1 0 -1 -2 -3 -4 1 0 -1 -2 -3 -4 -5 0 -1 -2 -3 -4 -5 -6 First-order ODEs Now, consider the first-order ODE dy dx = y − x. We just saw a table of values for the function y − x, so in a sense, we could ‘plot’ the values of the ODE above. For instance, dy dx x=−1,y =−2 = (−2) − (−1) = −1. But what does it mean that dy dx x=−1,y =−2 = −1? Remember that y (x) is a solution to the ODE, and that there are infinitely many solutions. So to say that dy dx x=−1,y =−2 = −1 means that any solution going through the point (−1, −2) must have a tangent line with slope −1. Vector Fields A vector field for a first-order ODE is a graphical representation, where at each point (x0 , y0 ) we draw a tiny line with slope dy dx x=x0 ,y =y0 to represent the slope of the tangent line at that point. Example Recall the table for f (x, y ) = y − x. Then, the vector field for dy dx = y − x is depicted below. Example Recall the table for f (x, y ) = y − x. Then, the vector field for dy dx = y − x is depicted below. Definitions An autonomous equation is a first-order ODE of the form dy dx = g (y ), i.e. the independent variable x does not appear (explicitly) in the ODE. Of course, y is still a function of x so must as well. dy dx An equilibrium solution of an autonomous equation dy dx = g (y ) is a solution of the form y = a, for some real number a. Finding Equilibrium Solutions ▶ To be an equilibrium solution means that you have the form y = a. d ▶ Then, dy dx = dx [a] = 0. ▶ Hence, finding equilibrium solutions amounts to solving the equation dy dx = 0. ▶ That is why these equilbrium solutions are also referred to as critical points for the ODE. Example Find the equilibrium solutions to the ODE. dy = y2 − y dx A Plot of the Equilibrium Solutions ODE: dy dx = y2 − y y =1 y =0 Solution: y (x) = 1 1−Ce x A Plot of the Equilibrium Solutions ODE: dy dx = y2 − y Solution: y (x) = 1 1+e x y =1 y (0) = 0.5 y =0 A Plot of the Equilibrium Solutions ODE: dy dx = y2 − y Solution: y (x) = 1 1−0.5e x y (0) = 2 y =1 y =0 A Plot of the Equilibrium Solutions ODE: dy dx = y2 − y Solution: y (x) = 1 1−1.5e x y =1 y =0 y (0) = −2 Definition Consider the first-order ODE dy dx = f (x, y ) and the corresponding IVP dy = f (x, y ); y (x ) = y . 0 0 An equilibrium solution y = a is dx said to be a stable solution if the following conditions are satisfied. Let y = g (x) be the solution to the IVP. ▶ If y0 > a, then g (x) converges down to y = a as x gets large. ▶ If y0 < a, then g (x) converges up to y = a as x gets large. Example ODE: dy dx =1−y Solution: y (x) = 1 − Ce −x Example ODE: dy dx =1−y Solution: y (x) = 1 − Ce −x Definition Consider the first-order ODE dy dx = f (x, y ) and the corresponding dy IVP dx = f (x, y ); y (x0 ) = y0 . An equilibrium solution y = a is said to be an unstable solution if the following conditions are satisfied. Let y = g (x) be the solution to the IVP. ▶ If y0 > a, then g (x) diverges up away from y = a as x gets large. ▶ If y0 < a, then g (x) diverges down away from y = a as x gets large. Example ODE: dy dx =y −1 Solution: y (x) = 1 + Ce x Example ODE: dy dx =y −1 Solution: y (x) = 1 + Ce x Definition Consider the first-order ODE dy dx = f (x, y ) and the corresponding IVP dy = f (x, y ); y (x ) = y . 0 0 An equilibrium solution y = a is dx said to be a semistable solution if it is neither stable nor unstable. Thus, solutions either converge to the equilibrium solution from above but diverge away from the equilibrium solution from below, or solutions converge from below but diverge from above. Example ODE: dy dx = (y − 1)2 Solution: y (x) = 1 + 1 C −x Example ODE: dy dx = (y − 1)2 Solution: y (x) = 1 + 1 C −x Determining the Stability of an Equilibrium Solution ▶ Find the equilibrium solutions (i.e. critical points) to the autonomous equation dy dx = g (y ). ▶ Create a number line for dy dx = g (y ), and place your critical points on the number line. ▶ Plug in numbers to determine whether dy dx is positive in each interval. ▶ For a critical point, y = a: if the number line shows dy dx is positive to the left of a critical point and negative to the right of a critical point, then y = a is a stable equilibrium solution. ▶ For a critical point, y = a: if the number line shows dy dx is negative to the left of a critical point and positive to the right of a critical point, then y = a is an unstable equilibrium solution. ▶ For a critical point, y = a: if the number line shows that dy dx does not change signs across a critical point, then y = a is a semistable equilibrium solution. Example Determine the stability of the equilibrium solutions to the following autonomous equation. dy = y 3 − 3y 2 dx Example Using our knowledge from the previous problem, which of the following functions (the red, brown, green, or teal) could possibly be a solution to the autonomous equation?